Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$. Let $F = ax^2 + bxy + cy^2 \in \mathfrak{F}$. We say $D = b^2 - 4ac$ is the discriminant of $F$. It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
Let $\Gamma =SL_2(\mathbb{Z})$.
Let $\alpha = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $\Gamma$.
We write $f^\alpha(x, y) = f(px + qy, rx + sy)$.
Since $(f^\alpha)^\beta$ = $f^{\alpha\beta}$, $\Gamma$ acts on $\mathfrak{F}$ from right.
Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). We denote the set of binary quadratic forms of discriminant $D$ by $\mathfrak{F}(D)$. It is easy to see that $\Gamma$ acts on $\mathfrak{F}(D)$ from right. Let $\sigma = \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right) \in \Gamma$. Let $\Gamma_{\infty} = \{\sigma^n; n \in \mathbb{Z}\}$. $\Gamma_{\infty}$ also acts on $\mathfrak{F}(D)$ from right. We denote the set of $\Gamma_{\infty}$-orbits on $\mathfrak{F}(D)$ by $\mathfrak{F}(D)/\Gamma_{\infty}$.
Let $K$ be a quadratic number field. Let $R$ be an order of $K$. Let $D$ be its discriminant. It is easy to see that $D$ is not not a square integer and $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
We denote the group of invertible elements of the ring $\mathbb{Z}$ by $\mathbb{Z}^\times$. Namely $\mathbb{Z}^\times = \{-1, 1\}$.
Let $F = ax^2 + bxy + cy^2 \in \mathfrak{F}(D)$. By this question, $I = \mathbb{Z}a + \mathbb{Z}\frac{(-b + \sqrt{D})}{2}$ is an ideal of $R$. Hence we get a map $\psi_0\colon \mathfrak{F}(D) \rightarrow \mathfrak{i}(R)$, where $\mathfrak{i}(R)$ is the set of fractional ideals of $R$. We define a map $\psi_1\colon \mathfrak{F}(D) \rightarrow \mathfrak{i}(R)\times\mathbb{Z}^\times$ by $\psi_1(F) = (\psi_0(F), sgn(a))$, where $sgn(a)$ is the sign of $a$. It is easy to see that $\psi_1$ induces a map $\mathfrak{F}(D)/\Gamma_{\infty} \rightarrow \mathfrak{i}(R)\times\mathbb{Z}^\times$. Hence this map induces a map $$\psi\colon\mathfrak{F}(D)/\Gamma_{\infty} \rightarrow (\mathfrak{i}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times$$
Next we would like to define a map $(\mathfrak{i}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times \rightarrow \mathfrak{F}(D)/\Gamma_{\infty}$
Let $I \in \mathfrak{i}(R)$. $I$ can be written as $I = J/\lambda$, where $J$ is an ideal of $R$ and $\lambda \in R$. We define the norm of $I$ as $N(I) = N(J)/N(\lambda R)$. It is easy to see that this is well defined.
Let $\alpha, \beta \in K$. We denote $\alpha\beta' - \alpha'\beta$ by $\Delta(\alpha, \beta)$, where $\alpha'$(resp. $\beta'$) is the conjugate of $\alpha$(resp. $\beta$). $\Delta(\alpha, \beta) \neq 0$ if and only if $\alpha, \beta$ are linearly independent over $\mathbb{Q}$. If $D < 0$, we define $\sqrt{D}$ as i$\sqrt{|D|}$. Let $\{\alpha, \beta\}$ be $\mathbb{Z}$-basis of $I \in \mathfrak{i}(R)$. If $\Delta(-\alpha, \beta)/\sqrt{D} > 0$, we say the basis $\{\alpha, \beta\}$ is positively oriented. If $\Delta(-\alpha, \beta)/\sqrt{D} < 0$, we say the basis $\{\alpha, \beta\}$ is negatively oriented.
Let $\{\alpha, \beta\}$ be positively oriented basis of $I \in \mathfrak{i}(R)$. We can assume that $\alpha \in \mathbb{Q}$. Let $x, y$ be indeterminates. Let $s \in \mathbb{Z}^\times$. We write $f(\alpha, \beta, s; x, y) = sN_{K/\mathbb{Q}}(x\alpha - sy\beta)/N(I)$. Namely $f(\alpha, \beta, s; x, y) = s(x\alpha - sy\beta)(x\alpha' - sy\beta')/N(I)$. It is easy to see that $f(\alpha, \beta, s; x, y)$ is a binary quadratic form of discriminant $D$.
My question Are the following propositions true? If yes, how do we prove them?
Proposition 1 The class of $\mathfrak{F}(D)/\Gamma_{\infty}$ represented by $f(\alpha, \beta, s; x, y)$ is determined only by the class of $(\mathfrak{i}(R)/\mathbb{Q}^\times)$ represented by $I$ and $s$.
Proposition 2 By proposition 1, we can define a map $$\phi\colon (\mathfrak{i}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times \rightarrow \mathfrak{F}(D)/\Gamma_{\infty}$$ by $\phi(([I], s)) = [f(\alpha, \beta, s; x, y)]$, where $[I]$(resp. $[f(\alpha, \beta, s; x, y)]$) denotes the class represented by $I$(resp. $f(\alpha, \beta, s; x, y)$).
Then $\psi$ and $\phi$ are inverses of each other.
Corollary Let $\mathfrak{F}_0(D)$ be the set of primitive binary quadratic forms of discriminant $D$. Let $\mathfrak{I}(R)$ be the group of invertible fractional ideals of $R$. Then the map $\psi$ induces a bijection:
$$\mathfrak{F}_0(D)/\Gamma_{\infty} \rightarrow (\mathfrak{I}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times$$
Proof: This follows immediately from proposition 2 and the proposition of this question.
